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1 Eigenvalues and Eigenvectors - Calvin University

1 eigenvalues and eigenvectors The product Ax of a matrix A Mn n (R) and an n-vector x is itself an n-vector. Of particular interest in many settings (of which differential equations is one) is the following question: For a given matrix A, what are the vectors x for which the product Ax is a scalar multiple of x? That is, what vectors x satisfy the equation Ax = x for some scalar ? It should immediately be clear that, no matter what A and are, the vector x = 0 (that is, the vector whose elements are all zero) satisfies this equation. With such a trivial answer, we might ask the question again in another way: For a given matrix A, what are the nonzero vectors x that satisfy the equation Ax = x for some scalar ?

1 = (1,0,3) and u 2 = (1,1,3). It is a fact that all other eigenvectors associated with λ 2 = −2 are in the span of these two; that is, all others can be written as linear combinations c 1u 1 +c 2u 2 using an appropriate choices of the constants c 1 and c 2. Example: Find the eigenvalues and associated eigenvectors of the matrix A = −1 2 0 ...

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1 Basic ANOVA concepts - Calvin University

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the table are the F values, and the P-values (the probability of getting an F statistic larger than that if the null hypothesis is true) are down the left side. Example: Determine P(F3,6 > 9.78)= 0.01 P(F2,20 > 5)= between 0.01 and 0.025 Example: A rm wishes to compare four programs for training workers to perform a certain manual task.

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[4] Computing Eigenvectors [5] Computing Eigenvalues [1] Eigenvectors and Eigenvalues Example from Di erential Equations Consider the system of rst order, linear ODEs. dy 1 dt = 5y 1 + 2y 2 dy 2 dt = 2y 1 + 5y 2 We can write this using the companion matrix form: y0 1 y0 2 = 5 2 2 5 y 1 y 2 : Note that this matrix is symmetric. Using notation ...

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0 3 1 0 0 3 1 A: Show that A is not diagonalizable. Solution: Use Theorem 5.2.2 and show that A does not have 3 linearly independent eigenvectors. I To nd the eigenvalues, we solve det( I A) = 1 1 1 0 + 3 1 0 0 + 3 = ( 1)( +3)2 = 0: So, = 1; 3 are the only eigenvalues of A: Satya Mandal, KU Eigenvalues and Eigenvectors x5.2 Diagonalization

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P is symmetric, so its eigenvectors .1;1/ and .1; 1/ are perpendicular. The only eigenvalues of a projection matrix are 0 and 1. The eigenvectors for D 0 (which means Px D 0x/ fill up the nullspace. The eigenvectors for D 1 (which means Px D x/ fill up the column space. The nullspace is projected to zero. The column space projects onto itself.

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Eigenvalues and Eigenvectors - MIT Mathematics

math.mit.edu

P is symmetric, so its eigenvectors .1;1/ and .1; 1/ are perpendicular. The only eigenvalues of a projection matrix are 0 and 1. The eigenvectors for D 0 (which means Px D 0x/ fill up the nullspace. The eigenvectors for D 1 (which means Px D x/ fill up the column space. The nullspace is projected to zero. The column space projects onto itself.

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If all 3eigenvalues are distinct →-−%≠0 Hence, /1"=0, i.e., the eigenvectors are orthogonal (linearly independent), and consequently the matrix !is diagonalizable. Note that a diagonalizable matrix !does not guarantee 3distinct eigenvalues.

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© 2012 Pearson Education, Inc. Slide 5.1- 10 EIGENVECTORS AND EIGENVALUES ! The scalar λ is an eigenvalue of A if and only if the equation has a nontrivial solution,

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Find all of the eigenvalues and eigenvectors of A= 1 1 0 1 : The characteristic polynomial is ( 1)2, so we have a single eigenvalue = 1 with algebraic multiplicity 2. The matrix A I= 0 1 0 0 has a one-dimensional null space spanned by the vector (1;0). Thus, the geometric multiplicity of this eigenvalue is 1.

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λ2 − 1 = 0, (12) which yields the eigenvalues, λ = ±1. The interpretation of this result is immediate. The matrix R(θ) when operating on a vector ~v represents a reflection of that vector through a line of reflection that passes through the origin. In the case of λ = 1 we have R(θ)~v = ~v, which means that ~v is a

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form, so we can read o the three eigenvalues: 1 = 1, 2 = 3, and 3 = 2. (It doesn’t matter the order you name them.) Thus, the spectrum of this matrix is the set f1;2;3g. Let’s nd the 1-eigenspace. We need to solve Ax = 1x. That’s the same as solving (A 1I)x = 0. The matrix A 1Iis 2 4 0 0 0 3 2 0 3 2 1 3 5 which row reduces to 2 4 1 0 1 6 ...

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